Problem 31 - Precalculus: Functions and Graphs

Question

Find all solutions of the equation. $$(2 \sin \theta+1)(2 \cos \theta+3)=0$$

Step-by-Step Solution

Answer

The solutions are \(\theta = \frac{7\pi}{6} + 2\pi n\) and \(\theta = \frac{11\pi}{6} + 2\pi n\) where \(n\) is an integer.

Step 1: Understand the Equation

We are given the equation \((2 \sin \theta+1)(2 \cos \theta+3)=0\). This is a factored equation set to zero, which means we need to find when either \(2 \sin \theta + 1 = 0\) or \(2 \cos \theta + 3 = 0\).

Step 2: Solve the First Equation

Solve \(2 \sin \theta + 1 = 0\). Subtract 1 from both sides to get \(2 \sin \theta = -1\). Divide by 2 to solve for \(\sin \theta\) giving us \(\sin \theta = -\frac{1}{2}\).

Step 3: Find Solutions for \(\sin \theta = -\frac{1}{2}\)

The sine function equals \(-\frac{1}{2}\) at angles \(\theta = \frac{7\pi}{6}\) and \(\theta = \frac{11\pi}{6}\), considering the unit circle, where sine is negative in the third and fourth quadrants.

Step 4: Solve the Second Equation

Solve \(2 \cos \theta + 3 = 0\). Subtract 3 from both sides to get \(2 \cos \theta = -3\). Divide by 2 to solve for \(\cos \theta\) giving us \(\cos \theta = -\frac{3}{2}\).

Step 5: Determine Validity of \(\cos \theta = -\frac{3}{2}\)

The cosine function ranges from \(-1\) to \(1\). Since \(-\frac{3}{2}\) is outside this range, there are no valid \(\theta\) for which \(\cos \theta = -\frac{3}{2}\). Thus, no solutions come from this equation.

Step 6: Combine Solutions

The solutions from the first equation are the only valid solutions. Therefore, the complete set of solutions for \(\theta\) are \(\theta = \frac{7\pi}{6}\) and \(\theta = \frac{11\pi}{6}\). If considering all solutions over the real numbers, add \(2\pi n\) where \(n\) is an integer.

Key Concepts

Sine Function

The sine function, denoted as \( \sin \theta \), is a fundamental component of trigonometry. It relates an angle in a right triangle to the ratio of the length of the side opposite the angle to the hypotenuse.

• The sine function is periodic, meaning it repeats its values in regular intervals.
• The period of the sine function is \(2\pi\), which means every \(2\pi\) radians, the function returns to its starting value.
• Sine values range from \(-1\) to \(1\).

In our problem, when solving for \( \sin \theta = -\frac{1}{2} \), we found the solutions using these properties. Specifically, understanding where on the unit circle the sine function takes the value of \(-\frac{1}{2} \), namely in the third and fourth quadrants, helped us find the precise angle solutions.

Cosine Function

The cosine function, \( \cos \theta \), is another key trigonometric function. It describes the ratio of the adjacent side to the hypotenuse in a right triangle.

• Like the sine function, cosine is also periodic with a period of \(2\pi\).
• The range of cosine is also from \(-1\) to \(1\).

In this task, we encountered the equation \( 2 \cos \theta + 3 = 0 \). By rearranging it, we can see that it attempts to set \(\cos \theta = \frac{-3}{2}\). However, since \(\frac{-3}{2}\) is not within the valid range of the cosine function, we know no solutions exist for this part.

Unit Circle

The unit circle is a powerful tool in trigonometry. It is a circle with a radius of one, centered at the origin of the coordinate plane. It is used to define trigonometric functions.

• The x-coordinate of a point on the unit circle is equal to \( \cos \theta \).
• The y-coordinate is equal to \( \sin \theta \).

For \( \theta = \frac{7\pi}{6} \) and \( \theta = \frac{11\pi}{6} \), the unit circle helps us determine that these angles correspond to the positions where the sine value is \(-\frac{1}{2}\). This match occurs in the third and fourth quadrants, where sine values are negative.

Quadrants

Quadrants divide the coordinate plane into four sections, each representing different signs of sine and cosine values:

• The first quadrant: both sine and cosine are positive.
• The second quadrant: sine is positive, but cosine is negative.
• The third quadrant: both sine and cosine are negative.
• The fourth quadrant: sine is negative, cosine is positive.

In this exercise, understanding the quadrants is essential because it allows us to determine where on the unit circle the sine function reaches \(-\frac{1}{2}\). These occur in the third and fourth quadrants, leading us to the solutions at \( \frac{7\pi}{6} \) and \( \frac{11\pi}{6} \).

Periodic Solutions

Trigonometric functions like sine and cosine are periodic, meaning they repeat at regular intervals over their domain.

• The period of these functions is \(2\pi\), so they return to the same value every \(2\pi\) radians.
• When addressing all solutions to an equation involving sine or cosine, you account for this periodicity.

In the exercise, the primary solutions for \( \sin \theta = -\frac{1}{2} \) are \( \theta = \frac{7\pi}{6} \) and \( \theta = \frac{11\pi}{6} \). To find all solutions, add \(2\pi n\) (where \(n\) is an integer) to these angles. This consideration gives an infinite set of solutions, each representing another cycle of the unit circle's rotation.